Question

quadratic polynomial whose zeroes 2,-1​

Answer 1

Step-by-step explanation:

The quadratic polynomial whose zeroes are $\frac{1}{2}$ and 1​ is $x^{2}$ - ($\frac{3}2}$)x + $\frac{1}{2}$ .

Let α and β be the roots of  are zeroes.

Here, α = $\frac{1}{2}$ and β = 1

To find, the quadratic polynomial whose zeroes are $\frac{1}{2}$ and 1​ = ?

We know that,

The equation of quadratic polynomial:

$x^{2}$ - (sum of zeros)x + Product  of zeroes

=$x^{2}$- (α + β)x + αβ

=$x^{2}$- ( $\frac{1}{2}$ + 1)x$\frac{1}{2}$ + . 1

=$x^{2}$- ($\frac{3}{2}$)x + $\frac{1}{2}$

Thus, the quadratic polynomial whose zeroes are $\frac{1}{2}$ and 1​ is $x^{2}$ - ($\frac{3}{2}$)x + $\frac{1}{2}$.

or

Answer:

2x^2-x+2

Step-by-step explanation:

let a+b=1/2

& a×b=1

now,

x^2-(sum of zeros)x+product

x^2-(a+b)x+a×b

×2-1/2x+1

by taking LCM

2X^2-x+2